Question

The general form of the complementary function of a differential equation is given by: \[ y(t) = (At + B)e^{-2t}, \] where \(A\) and \(B\) are real constants determined by the initial condition. The corresponding differential equation is:

• A. \(\frac{d^2y}{dt^2} + 4 \frac{dy}{dt} + 4y = f(t)\)
• B. \(\frac{d^2y}{dt^2}+ 4y = f(t)\)
• C. \(\frac{d^2y}{dt^2} + 3 \frac{dy}{dt} + 2y = f(t)\)
• D. \(\frac{d^2y}{dt^2} + 5 \frac{dy}{dt} + 6y = f(t)\)

Hint:

If the complementary function includes terms like \((At + B)e^{rt}\), it signifies repeated roots in the characteristic equation, leading to additional polynomial factors.

Answer 1

The Correct Option is A

Step 1: Analyze the complementary function.
The given complementary function is: \[ y_c(t) = (At + B)e^{-2t}. \] This implies that the characteristic equation has a repeated root at \(-2\). Step 2: Formulate the characteristic equation.
The complementary solution corresponds to the characteristic equation: \[ (r + 2)^2 = 0 \quad \Rightarrow \quad r = -2, \, -2. \] Step 3: Determine the differential equation.
The characteristic equation \((r + 2)^2 = 0\) translates to the following differential equation: \[ \frac{d^2y}{dt^2} + 4 \frac{dy}{dt} + 4y = f(t). \] Final Answer: \[\boxed{{(1) \(\frac{d^2y}{dt^2} + 4 \frac{dy}{dt} + 4y = f(t)\)}}\]